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enter image description here I'm coming across a problem that states $\lim_{x\to c}f(c)=f(c)$. But I'm really confused by this. What does $\lim_{x\to c}f(c)=f(c)$ even mean?

For instance let's say $f(x)=x^2$, and $c=2$. So $\lim_{x\to c}f(c)$ means $\lim_{x\to 2}4$?. What does that even mean? $4$ isn't a function, it's just a number, how can you find the limit of it?

Can someone please explain this to me? Try to avoid using epsilon delta proofs - I'm still just learning Khan Academy Calculus.

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    $\begingroup$ $f(c)=4$ can simply be viewed as a constant function, so no matter what value $x$ takes, however close you may bring $x$ to $2$, the function's value is still $4$, and by definition, it's limit as $x\to 2$ remains $4$ $\endgroup$
    – GoodDeeds
    Commented Sep 2, 2018 at 11:52
  • $\begingroup$ @GoodDeeds I included a screenshot above, it's from a proof that all differentiable functions are continuous. You say it's a constant function. But then how is it a function when it is just and individual number? Thanks by the way. $\endgroup$
    – Ethan Chan
    Commented Sep 2, 2018 at 11:56
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    $\begingroup$ An individual number is a constant function. Its value does not depend on $x$. $\endgroup$
    – GoodDeeds
    Commented Sep 2, 2018 at 11:57
  • $\begingroup$ Instead of using limit properties with respect to subtraction, consider the limit of the difference, without rewriting as two separate limits. When $x = c + 0.1,$ then $f(x) = f(c + 0.1) - f(c).$ When $x = c + 0.01,$ then $f(x) = f(c + 0.01) - f(c).$ When $x = c + 0.001,$ then $f(x) = f(c + 0.001) - f(c).$ When $x = c + 0.0001,$ then $f(x) = f(c + 0.0001) - f(c).$ Do you see that the values of $f(x)$ are approaching $\;\lim_{x \rightarrow c} f(x) \;\; - \;\; f(c)?$ $\endgroup$
    – Dave L. Renfro
    Commented Sep 2, 2018 at 12:16

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It means that as $x$ comes close to $c$, the value of $f(c)$ comes close to, well, $f(c)$. Because $f(c)$ doesn't care what $x$ is.

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